CONTENTS
Page
1.
Kinematics of circular motion
39
2.
Uniform circular motion and
41
Non-uniform circular motion
3.
47
Dynamics of circular motion
N
(Circular turning on roads, Conical
0
pendulum, Death well or Rotor)
4.
Vertical circular motion
5.
Exercise-I (Conceptual Questions)
6.
Exercise-II (Previous Years Questions)
61
7.
Exercise-III (Analytical Questions)
63
8.
Exercise-IV (Assertion & Reason)
64
19
-2
49
55
E
Se
A
ss
io
n
20
LL
E
CIRCULAR MOTION
S.No.
NEET SYLLABUS
Uniform circular motion, Dynamics of uniform circular motion. Centripetal force, examples of circular motion (vehicle
on level circular road, vehicle on banked road). Vertical circular motion.
RAJA RAMANNA
He was born on January 28, 1925 at Tumkur, Karnataka. hand picked
by the founder of India’s nuclear program, Dr. Homi Bhabha, Dr. Raja
Ramanna was a celebrated physicist and nuclear scientist that India had
ever produced. A multifaceted personality Dr. Raja Ramanna played
the roles of a technologist, nuclear physicist, administrator, leader,
musician, Sanskrit literature scholar, and philosophy researcher. To
N
complete the endless list of honors that this nobleman was gifted with,
he was a complete human being. Following the steps of his ideals Dr.
Homi Bhabha and Vikram Sarabhai, Ramanna managed to grab a major
0
position in shaping india’s energy and security programs. He is regarded as one of the most successful
19
20
LL
E
experiment.
-2
creators of science and technology in India with the tremendous success of India’s peaceful explosion
VIKRAM SARABHAI
n
Vikram Ambalal Sarabhai was born on August 12, 1919 at Ahmedabad
ss
io
in an affluent family of progressive industrialists. He was considered the
father of the Indian space program ; instrumental in establishing the
physical research laboratory (PRL) in Ahmedabad in November 1947;
was Chairman of the Atomic Energy Commission. Apart from being a
Se
scientist, he was a rare combination of an innovator, industralist and
A
visionary. Vikram Sarabhai did research on the time variations of cosmic
rays and concluded that meteorological effects could not entirely affect
the observed daily variations of cosmic rays ; further, the residual variations
were wide and global and these were related to variations in solar activity.
Vikram Sarabhai visualized a new field of research opening up in solar and interplanetary Physics. Sarabhai
set up the first Rocket Launching station (TERLS) in the country at Thumba near Thiruvananthapuram on
the Arabian Coast, as Thumba is very close to the Equator. The first rocket with sodium vapour payload was
launched on November 21, 1963. In 1965, the UN General Assembly gave recognition to TERLS as an
international facility.
Pre-Medical : Physics
ALLEN
CIRCULAR MOTION
If a particle moves in a plane such that its distance from a fixed (or moving) point remains constant then
its motion is called as circular motion with respect to that fixed or moving point.
That fixed point is called the centre and the corresponding distance is called the radius of circular path.
The vector joining the centre of the circle and the particle performing circular motion, directed towards the
later is called the radius vector. It has constant magnitude but variable direction.
1.
KINEMATICS OF CIRCULAR MOTION
Angular Displacement
Angle traced by the position vector of a particle moving w.r.t. some fixed point is called angular displacement.
Q
P
Q
angle =
\
Dq =
arc PQ
r
0
Dq = angular displacement
-2
Dq
r
arc
radius
N
fixed
point
Note : 1 rps = 60 rpm
19
LL
E
Frequency (n) : Number of revolutions described by particle per second is its frequency. Its unit is revolutions
per second (rps) or revolutions per minute (rpm).
20
Time Period (T) : It is the time taken by particle to complete one revolution. i.e. T =
1
n
Þ
angle traced
Dq dq
= Lim
=
D
t
®
0
time taken
Dt
dt
Its unit is rad/s and dimensions is [T–1]
ss
io
w=
n
Angular Velocity (w) : It is defined as the rate of change of angular displacement of a moving particle,
w.r.t. to time.
arc
Ds
=
radius
r
A
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Angle (Dq) =
E
\
Þ
Se
Relation between linear and Angular velocity
Ds = rDq
Here,
Dq
Ds
r
P
ds
dq
Ds
rDq
=
=r
if Dt ® 0 then
Dt
Dt
dt
dt
In vector form
Q
r
Þ
v = wr
r r r
v = w ´ r (direction of vr is according to right hand thumb rule)
r
v = Linear velocity
[Tangential vector]
r
w = Angular velocity
[Axial vector]
r = Radius vector or position vector
r
Note : Centrifugal means away from the centre and centripetal means towards the centre.
39
Pre-Medical : Physics
ALLEN
r
r r
All the three vectors v, w and r are mutually perpendicular to each other..
r
Here, vr ^ w ^ rr Q
r r
v^w
\
rr
v.w = 0
Q
r r
w^r
\
rr
w.r = 0
r r
v^r
\
rr
v.r = 0
Q
w
v
r
Average Angular Velocity (wav)
wav =
t = t1
total angle of rotation
Dq
q - q1
=
= 2
total time taken
Dt
t2 - t1
q2
t = t2
q1
t=0
where q1 and q2 are the angular positions of the particle at instants t1 and t2.
Instantaneous Angular Velocity
Angular Acceleration (a)
v A sin q1 + v B sin q2
r
ss
io
Here (vAB)^ = vA sin q1 + vB sin q2 \ wAB =
20
relative velocity of A w.r.t. B perpendicular to line AB
(v AB ) ^
=
separation between A and B
rAB
n
wAB =
r
19
LL
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B is the angular velocity of the position vector of A w.r.t. B. It means
the rate at which the position vector of 'A' w.r.t. B rotates at that instant.
-2
Relative angular velocity of a particle 'A' w.r.t. an other moving particle
0
Relative Angular Velocity
N
r
Dq dq
r
Lim
=
The angular velocity at some particular instant w = Dt ®0
Dt dt
Average Angular Acceleration :
Se
r
r
r
Dw dw
=
Rate of change of angular velocity is called angular acceleration. i.e. a = Lim
Dt ® 0 D t
dt
Relation between Angular and Linear Accelerations
r r r
Velocity v = w ´ r
r
r
r
r dv
d r r
dw r r dr
´r+w´
(w ´ r) =
Acceleration a =
=
dt
dt
dt
dt
r r
r
r r r r r
Þ a = a ´ r + w ´ v = a = a T + aC
w
r
r r
r
r r
( a T = a ´ r is tangential acceleration & a C = w´ v is centripetal acceleration)
r r
r
r
r
a = a T + a C ( a T and a C are the two component of net linear acceleration)
r
r
r
As a T ^ a C so |a|= a2T + a2C
40
axis of rotation
v
r
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A
r
r
change in angular velocity Dw
a avg =
=
time taken
Dt
E
Pre-Medical : Physics
ALLEN
Tangential Acceleration
r
r
r r
r r
r
r
a T = a ´ r , its direction is parallel (or antiparallel) to velocity. v = w ´ r as w and a both are parallel
(or antiparallel) and along the axis.
Magnitude of tangential acceleration in case of circular motion :
r
r
r
r
aT = a r sin 90° = ar ( a is axial, r is radial so that a ^ r )
r
r
r
r
As a T is along the direction of motion (in the direction of v or opposite to v ) so a T is responsible for change
in speed of the particle. Its magnitude is the rate of change of speed of the particle.
Note : If a particle is moving on a circular path with constant speed then tangential acceleration is zero.
Centripetal acceleration
r r r
r
r r
r r r
a C = w ´ v = w ´ (w ´ r) (Q v = w ´ r )
axis of rotation
w
r
r
Let r be in î direction and w be in ĵ direction then the direction of
O
r
P
N
r
r
ˆ = – $i , opposite to the direction of r
a C is along ˆj ´ (ˆj ´ ˆi) = ĵ ´ ( -k)
2.
UNIFORM AND NON-UNIFORM CIRCULAR MOTION
20
2.1 Uniform Circular Motion
19
Note : Centripetal acceleration is always perpendicular to the velocity at each point.
-2
v2
r
v2
= w 2 r therefore a C =
( -ˆr)
r
r
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acceleration, a C = wv =
0
i.e., from P to O and it is centripetal in direction. Magnitude of centripetal
n
If a particle moves with a constant speed in a circle, the motion is called uniform circular motion. In uniform
circular motion a resultant non-zero force acts on the particle. The acceleration is due to the change in
direction of the velocity vector. In uniform circular motion tangential acceleration (aT) is zero. The acceleration
v2
. Here, v is the speed of the particle and r the
r
ss
io
of the particle is towards the centre and its magnitude is
radius of the circle.
A
(as v = rw )
Se
The direction of the resultant force F is therefore, towards the centre and its magnitude is F =
mv 2
=mrw2
r
Here, w is the angular speed of the particle. This force F is called the centripetal force. Thus, a centripetal
mv 2
is needed to keep the particle moving in a circle with constant speed. This force
r
is provided by some external agent such as friction, magnetic force, coulomb force, gravitational force,
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force of magnitude
E
tension. etc.
In this motion :
l
Speed = constant
l
Velocity ¹ constant (becuase its direction continuously changes)
l
K.E. =
l
aT = 0
1
mv2 = constant
2
Velocity = constant
l
l
r
w = constant (because magnititude and direction, both are constants)
é
ù
dv d ( constant )
= 0ú
êëQ a T = dt =
dt
û
41
Pre-Medical : Physics
l
l
aT 0
é
ù
êëQ a = r = r = 0úû
a= 0
or
dw d
é
ù
êë a = dt = dt ( constant ) = 0 úû
r r
r r
v2
a = a cp ¹ constant
a = a cp = wv = w2r =
= constant
l
r
(because the direction of acp is toward the centre of circle which changes as the particle revolves)
r r
Total work done = Fnet .s
r
r
éQ Fnet = Fcp ù
ë
û
r r
= Fcp .s
90°
= Fcp s cos90° = 0
Work 0
= = 0 or
time
t
O
Fcp
r r r r
Power = Fnet .v = Fcp .v = Fcpvcos90° = 0
Power =
l
Uniform circular motion is usually executed in a horizontal plane.
N
l
Example :
A particle of mass 'm' is tied at one end of a string of length
T=
mv 2
r
-2
mg
20
So,
T
19
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'r' and it is made to revolve along a circular path in a horizontal
plane with a constant speed means a (uniform circular
motion) In this condition the required centripetal force is
provided by the tension in the string.
T = Fcp
v
0
l
ALLEN
aT
n
2.2 Non-Uniform Circular Motion :
If a particle moves with variable speed in a circle, then the motion is called nonuniform circular motion.
Accleration (a) has two components :-
acp
Se
r
a cp = responsible for change in direction only.
ss
io
In this motion :
a
A
r
a T = responsible for change in speed only.
speed = |velocity|is variable,
As speed variable hence given physical quantities are also variable.
r
v
w =
r
l
K.E. =
1
mv2
2
l
l
r
v2
a cp = wv = w2 r =
r
l
r r
r
a = a T + a cp
l
r
2
F = FT2 + Fcp
l
42
r r
r
F = FT + Fcp
l
a¹ 0
l
æ v2 ö
r
2
a = a T2 + acp2 = ( ar ) + ç ÷
è r ø
l
aT ¹ 0
2
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Hence due to aT
E
Pre-Medical : Physics
ALLEN
l
l
Work done by centripetal force will be
FT
zero but work done by tangential force is not zero.
s
Total work done
90°
Fcp
r r
W = FT .s = FT s cos 0° = FTs
Þ
l
l
(where s is the distance travelled by the particle)
r r
work FT .s r r
=
= FT .v = FTv cos 0° = FTv
Power =
time
t
a
Angle between velocity and acceleration is given by :
tan q =
l
v
acp
aT
=
aT
Fcp
q
FT
acp
Example
N
Circular motion in vertical plane is an example of non-uniform circular motion.
2.3 Equations of circular motion
l
Final velocity (v)
Final angular velocity (w)
l
Displacement (s)
Angular displacement (q)
l
Acceleration (a)
Angular Acceleration (a)
If
1 2
at
2
q = w0t +
w2 = w02 + 2aq
ss
io
v2 = u2 + 2as
a
(2n–1)
2
qnth = w0 +
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A
æu+vö
s =ç
÷t
è 2 ø
a
(2n –1)
2
æ w + wö
q =ç 0
÷t
è 2 ø
r
Se
sn th = u +
1 2
at
2
n
w = w0 + at
v = u + at
s = ut +
a = constant, then
20
a = constant, then
0
Initial angular velocity (w0)
-2
Initial velocity (u)
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l
If
E
Rotational Motion
19
Translatory / Linear Motion
GOLDEN KEY POINTS
•
Small angular displacement dq is a vector quantity, but large angular displacement q is not a vector quantity.
r r
r r
r
r
r
r
dq1 + dq2 = dq2 + dq1 But q1 + q2 ¹ q2 + q1
•
Direction of angular displacement is perpendicular to the plane of rotation and is given by right hand thumb
rule.
•
Angular displacement is dimensionless and its S.I. unit is radian while other units are degree and revolution.
2p radian = 360° =1 revolution
•
Instantaneous angular velocity is an axial vector quantity.
•
Direction of angular velocity is same as that of angular displacement i.e. perpendicular to the plane of rotation
and along the axis according to right hand screw rule or right hand thumb rule.
43
Pre-Medical : Physics
•
r
r
If particles A and B are moving with a velocity v A and v B
ALLEN
®
vA
®
vB
and are separated by a distance r at a given instant then
(i)
d qBA
dr
v B sin q2 - v A sin q1
= vBcosq2–vAcosq1 (ii)
= wBA =
dt
dt
r
q1
A
q2
r
B
•
Angular acceleration is an axial vector quantity. It's direction is along the axis according to the right hand
thumb rule or right hand screw rule.
•
Important difference between projectile motion and uniform circular motion :
In projectile motion, both the magnitude and the direction of acceleration (g) remain constant, while in uniform
circular motion the magnitude remains constant but the direction continuously changes.
Illustrations
Illustration 1.
N
A particle revolving in a circular path completes first one third of the circumference in 2 s, while next one
third in 1s. Calculate its average angular velocity.
Solution
wPQ =
8 sin 30o - ( -6 sin 30o )
10
= 0.7 rad/s.
0
ss
io
Solution
n
Velocity of P is 8 m/s and that of Q is 6 m/s at 30° angle with the
line joining P & Q. Calculate the angular velocity of P w.r.t. Q
20
Two moving particles P & Q are 10 m apart at any instant.
-2
Illustration 2.
rad/s
19
LL
E
2p 2p 4 p
+
q1 + q2
4p
2p
2p
3
3 = 3 =
q1 =
and q2 =
total time T = 2 + 1 = 3 s \ < wav > =
=
3
3
T
9
3
3
Illustration 3.
A particle is moving parallel to x–axis as shown in fig. such that the y component
Se
of its position vector is constant at all instants and is equal to 'b'. Find
makes an angle q with the x-axis.
Solution
b
\
wPO =
v sin q v 2
= sin q
b
b
sin q
Illustration 4.
The angular velocity of a particle is given by w =1.5t – 3t2 +2. (where t is in seconds). Find the instant when
its angular acceleration becomes zero.
Solution
a=
44
dw
= 1.5 – 6t = 0 Þ t = 0.25 s.
dt
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A
the angular velocity of the particle about the origin when its radius vector
E
Pre-Medical : Physics
ALLEN
Illustration 5.
A disc starts from rest and gains an angular acceleration given by a = 3t – t2 (where t is in seconds) upon
the application of a torque. Calculate its angular velocity after 2 s.
Solution
a=
dw
= 3t–t2
dt
w
t
0
0
2
Þ ò dw = ò (3t - t )dt Þ w =
3t 2 t 3
10
rad/s
Þ at t =2 s, w =
3
2
3
y
Illustration 6.
A particle is moving in clockwise direction in a circular path as shown in figure.
r
The instantaneous velocity of particle at a certain instant is v = (3$i + 3$j ) m/s.
II
I
III
IV
x
Then in which quardant does the particle lie at that instant ? Explain your answer.
Solution
0
x
IV
19
LL
E
III
I
-2
II
N
II quardant. According to following figure x & y components of velocity are positive when the particle is
in II quardant.
y
Illustration 7.
Centripetal acceleration ac =
Net acceleration (a) =
at t = 1s, aT = 4 m/s2
v 2 4t 4
=
= 4t4,
r
1
a 2T + a 2c =
n
dv
= 4t,
dt
ss
io
Tangential acceleration = aT =
at t = 1 s,
ac = 4 m/s2
2
42 + 42 = 4 2 m/s .
A
Illustration 8.
Se
Solution
20
A particle is performing circular motion of radius 1 m. Its speed is v = (2t 2) m/s. What will be the magnitude
of its acceleration at t = 1s ?
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A cyclist is riding with a speed of 18 km/h. As he approaches a circular turn on the road of radius 25 2 m,
E
he applies brakes which reduces his speed at a constant rate of 0.5 m/s every second. Determine the magnitude
and direction of his net acceleration on the circular turn.
Solution
®
v
5
dv
1
v2
25
1
= - m / s2
v = 18 ×
=5 m/s and acp=
=
=
m/s2 , aT = 18
dt
2
R 25 2
2
a CP
1 2
æ 1 ö æ1ö
3
=
= 0.86 m / s2 , tanq = a =
ç
÷ +ç ÷ =
T
12
2
è 2 ø è2ø
2
anet =
q = tan -1
2
e 2j from tangential direction
2
2
=
®
aCP
2
®
anet
®
a
T
45
Pre-Medical : Physics
ALLEN
Illustration 9.
®
aT
A particle is moving in a circular orbit with a constant tangential acceleration
starting from rest. After 2 s of the beginning of its motion, angle between
the acceleration vector and the radius becomes 45°. What is the angular
®
a
45°
®
aCP
acceleration of the particle ?
Solution
Center
r
r
In the adjoining figure the total acceleration vector a and its components-the tangential acceleration a T
r
and normal acceleration a CP are shown. These two components are always mutually perpendicular to each
other and act along the tangent to the circle and radius respectively. Therefore, if the total acceleration vector
makes an angle of 45° with the radius, both the tangential and the normal components must be equal in
magnitude.
aT = acp Þ aR = w2R Þ a = w2
N
...(i)
Since angular acceleration is uniform, we have w = wo + at
From equations (i) and (ii), we have
a = 0.25 rad/s2
...(ii)
(A) 8 rad/s2
(B) 10 rad/s2
(C) 12 rad/s2
(D) None of these
ss
io
n
The second's hand of a watch has 6 cm length. The speed of its tip and magnitude of difference in velocities
of its at any two perpendicular positions will be respectively :
(B) 2 2 p & 4.44 mm/s
(A) 2p & 0 mm/s
(C) 2 2 p & 2p mm/s
Se
(D) 2p & 2 2p mm/s
A particle is moving on a circular path of radius 6 m. Its linear speed is v = 2t, here t is time in second and
A
3.
20
If angular velocity of a particle depends on the angle rotated q as w = q2 + 2q, then its angular acceleration
a at q = 1 rad is :
2.
19
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BEGINNER'S BOX-1
1.
0
w =2a
-2
Substituting w0=0 and t = 2 s, we have
v is in m/s. Calculate its centripetal acceleration at t = 3 s.
Two particles move in concentric circles of radii r 1 and r2 such that they maintain a straight line through the
centre. Find the ratio of their angular velocities.
5.
If the radii of circular paths of two particles are in the ratio of 1 : 2, then in order to have same centripetal
acceleration, their speeds should be in the ratio of :
(A) 1 : 4
6.
(B) 4 : 1
(C) 1 : 2
(D)
2 :1
A stone tied to the end of a 80 cm long string is whirled in a horizontal circle with a constant speed. If the
stone makes 14 revolutions in 25 s, the magnitude of its acceleration is :
(A) 20 m/s2
46
(B) 12 m/s2
(C) 9.9 m/s2
(D) 8 m/s2
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4.
E
Pre-Medical : Physics
ALLEN
7.
For a body in a circular motion with a constant angular velocity, the magnitude of the average acceleration
over a period of half a revolution is.... times the magnitude of its instantaneous acceleration.
2
p
(B)
(C) p
(D) 2
p
2
A ring rotates about z axis as shown in figure. The plane of rotation is xy. At a certain instant the acceleration
(A)
8.
(
)
of a particle P (shown in figure) on the ring is 6iˆ - 8jˆ m / s2 . Find the angular acceleration of the ring and
its angular velocity at that instant. Radius of the ring is 2 m.
y
P
x
DYNAMICS OF CIRCULAR MOTION
3.1 Circular Turning on Roads
0
3.
N
O
-2
When vehicles go through turnings, they travel along a nearly circular arc. There must be some force which
LL
E
provides the required centripetal acceleration. If the vehicles travel in a horizontal circular path, this resultant
19
force is also horizontal. The necessary centripetal force is being provided to the vehicles by the following three
ways :
• By banking of roads only.
• By friction and banking of roads both.
20
• By friction only.
In real life the necessary centripetal force is provided by friction and banking of roads both.
•
By Friction only
n
Suppose a car of mass m is moving with a speed v in a horizontal circular arc of radius r. In this case, the
circular path.
Thus, f =
mv 2
r
ss
io
necessary centripetal force will be provided to the car by the force of friction f acting towards centre of the
Q fmax =mN = mmg
2
•
By Banking of Roads only
mv
mv 2
£ fmax Þ
£ mmg Þ
r
r
Se
A
Therefore, for a safe turn without skidding
v £ mrg .
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Friction is not always reliable at turns particularly when high speeds and sharp
E
N N cos q
turns are involved. To avoid dependence on friction, the roads are banked
q
at the turn in the sense that the outer part of the road is some what lifted
N sin q =
mv
r
N sin q
compared to the inner part.
mv 2
and N cos q = mg
r
v2
Þ tan q =
Þ v = rg tan q
rg
v2 h
=
Note : tanq =
rg b
2
h
h
Q tan q =
b
mg
q
b
47
Pre-Medical : Physics
•
ALLEN
Friction and Banking of Road both
If a vehicle is moving on a circular road which is rough and banked also, then three forces may act on the
vehicle, of these the first force, i.e., weight (mg) is fixed both in magnitude and direction. The direction of
second force, i.e., normal reaction N is also fixed (perpendicular to road) while the direction of the third force,
i.e., friction f can be either inwards or outwards while its magnitude can be varied upto a maximum limit (f max
= mN ).
So, direction and the magnitude of friction f are so adjusted that the resultant of the three forces mentioned
In this case, N cos q + f sin q = mg
q
N sin q
q
...(ii)
æ tan q - µ ö
Rg ç
è 1 + µ tan q ÷ø
20
vmin =
ss
io
æ tan q - tan f ö
Rg ç
=
è 1 + tan f tan q ÷ø
Rg tan(q – f)
If speed of vehicle is high then friction force act inwards.
A
in this case for maximum speed
Se
(b)
n
If we assume µ = tan f, then
vmin =
19
LL
E
Therefore
-2
For minimum speed f = µN so by dividing Eqn. (1) by Eqn. (2)
N cos q - µN sin q
mg
=
2
N sin q - µN cos q mv min / R
N cos q – µN sin q = mg
and
N sin q + µN cos q =
which gives vmax =
2
mv
r + F cos q
mg
N
and
...(i)
mv2
N sin q – f cos q =
R
N cos q + F sin q
f
q
N
If speed of the vehicle is small then friction acts outwards.
mv 2max
N
q
N sin q + F cos q
q
f
q
N cos q
mv
r
2
mg + F sin q
R
æ tan q + µ ö
Rg ç
è 1 - µ tan q ÷ø =
Rg tan(q + f )
Hence for successful turning on a rough banked road, velocity of vehicle must satisfy following relation
Rg tan(q - f ) £ v £ Rg tan(q + f)
where q = banking angle and f = tan–1(µ).
48
Z:\NODE02\B0AI-B0\TARGET\PHY\ENG\MODULE_02\02-CIRCULAR MOTION\01-THEORY.P65
(a)
mv 2
towards the centre.
r
0
above is
E
Pre-Medical : Physics
ALLEN
3.2 Conical Pendulum
If a small particle of mass m tied to a string is whirled along a horizontal circle, as shown in figure then the
arrangement is called a 'conical pendulum'. In case of conical pendulum the vertical component of tension balances
the weight while its horizontal component provides the necessary centripetal force. Thus,
T sin q =
mv 2
r
and T cos q = mg Þ v =
\ Angular speed
w=
rg tanq
q
L
g tan q
r
v
=
r
T
r
T sin q
r
L cos q
2p
So, the time period of pendulum is T =
= 2p
= 2p
g
g tan q
w
r=Lsinq
T cos q
q
m
mv
r
2
mg
3.3 'Death Well' or Rotor
In case of 'death well' a person drives a motorcycle on the vertical
N
surface of a large wooden well while in case of a rotor a person hangs
resting against the wall without any support from the bottom at a certain
angular speed of rotor. In death well walls are at rest and person revolves
while in case of rotor person is at rest and the walls rotate.
N
mg
Death well
VERTICAL CIRCULAR MOTION
20
4.
mv 2
2
= mrw
r
19
f = mg and N =
-2
LL
E
In both cases friction balances the weight of person while reaction
provides the centripetal force for circular motion, i.e.,
r
0
f
O
q
A
ss
io
n
Suppose a particle of mass m is attached to a light inextensible string of length R. The particle is moving
in a vertical circle of radius R about a fixed point O. It is imparted a velocity u in the horizontal direction
at lowest point A. Let v be its velocity at point P of the circle as shown in the figure.
T
P
q mgcosq
Se
mgsinq mg
Z:\NODE02\B0AI-B0\TARGET\PHY\ENG\MODULE_02\02-CIRCULAR MOTION\01-THEORY.P65
A
When a particle is whirled in a vertical circle then three cases are possible -
E
Case I :
Particle oscillates in lower half circle.
Case II :
Particle moves to upper half circle but not able to complete loop.
Case III : Particle completes loop.
(
Condition of Oscillation 0 < u £ 2gR
)
The particle will oscillate if velocity of the particle becomes zero but tension in the string is not zero.
(In lower half circle (A to B))
mv 2A
R
q
mv 2A
+ mgcosq
T=
R
A q
in
gs
m
Here, T – mgcosq =
B
T
vA
v
v=0
T¹0
q
mg
mg cosq
49
Pre-Medical : Physics
ALLEN
In the lower part of circle when velocity become zero and tension is non zero means when v = 0, but T ¹ 0
So, to make the particle ossillate in lower half cycle, maximum possible velocity at A can be. given by
1
mv 2A + 0 = mgR + 0
(by COME between A and B)
2
vA =
....(i)
2gR
Thus, for 0 < u £ 2gR , particle oscillates in lower half of the circle (00 < q £ 90°)
This situation is shown in the figure. 0 < u £ 2gR or 0°<q £ 90°
Condition of Leaving the Circle :
( 2gR < u < 5gR )
C
v
qT
B
N
q
æ mv 2
ö
T = çç R – mg cos q ÷÷
è
ø
20
.....(ii)
-2
mv2
R
LL
E
Here, T + mg cosq =
19
In upper half cycle (B to C)
0
mg
In this part of circle tension force can be zero without having zero velocity mean when T = 0, v ¹ 0
vc
ss
io
n
form equation (ii) it is clear that tension decreses if velocity decreases. So to complete the loop tension
force should not be zero, in between B to C. Tension will be minimum at C i.e., T c ³ 0 is the required
condition.
C
Tc + mg =
mv2c
R
vc =
gR
if Tc = 0
Then
mg =
mv2c
R
vc2 = gR
Þ
By COME (Between A and C)
1
1
mv 2A + 0 = mv 2c + mg(2R)
2
2
vA2 = vc2 + 4gR
Therefore, if
2gR < u <
Þ
vA2 = 5gR Þ v A = 5gR
5gR , the particle leaves the circle.
Note : After leaving the circle, the particle will follow a parabolic path.
50
Z:\NODE02\B0AI-B0\TARGET\PHY\ENG\MODULE_02\02-CIRCULAR MOTION\01-THEORY.P65
At Top
Se
A
mg
Tc
E
Pre-Medical : Physics
ALLEN
(
Condition of Looping the Loop u ≥ 5gR
)
vTop ³ Rg
TTop ³ 0
The particle will complete the circle if the string does not
slack even at the highest point ( q = p ) . Thus, tension in
the string should be greater than or equal to zero (T ³ 0)
at q=p. In critical case substituting T=0
vHz ³ 3Rg
THz ³ 3mg
vHz ³ 3Rg
THz ³ 3mg
Thus, if u ³ 5gR , the particle will complete the circle.
Note : In case of light rod tension at top most point can never
be zero so velocity will become zero.
\ For completing the loop
vL ³ 5Rg
TL ³ 6mg
v L ³ 4gR
Illustrations
Illustration 10.
120 » 11 m/s
19
mrg =
LL
E
mv 2
£ mmg Þ vmax =
r
Illustration 11.
0
Here centripetal force is provided by friction so
-2
Solution
N
Find the maximum speed at which a car can turn round a curve of 30 m radius on a levelled road if the
coefficient of friction between the tyres and the road is 0.4 [acceleration due to gravity = 10 m/s2]
So tan q =
5
50
v2
. Here v= 60 km/h = 60 ×
m/s =
m/s r = 0.1 km = 100 m
18
3
rg
50 / 3 ´ 50 / 3
5
=
Þ
18
100 ´ 10
Illustration 12.
n
In case of banking tan q =
ss
io
Solution
20
For traffic moving at 60 km/h, if the radius of the curve is 0.1 km, what is the correct banking angle of
the road ? (g = 10 m/s2)
æ 5ö
q = tan -1 ç ÷ .
è 18 ø
Se
A hemispherical bowl of radius R is rotating about its axis of symmetry which
Z:\NODE02\B0AI-B0\TARGET\PHY\ENG\MODULE_02\02-CIRCULAR MOTION\01-THEORY.P65
A
is kept vertical. A small ball kept in the bowl rotates with the bowl without slipping
on its surface. If the surface of the bowl is smooth and the angle made by the
2
radius through the ball with the vertical is a . Find the angular speed at which mw r
the bowl is rotating.
E
Solution
N cosa = mg
.....(1)
N sina = mrw2
.....(2)
r = Rsina
.....(3)
w
N cosa
aR N a
R
A
r
N sina
mg
From equations (2) & (3)
Nsina = mw2Rsina
N=mRw2
Þ (mRw2) cosa = mg
.....(4)
Þw=
g
R cos a
51
Pre-Medical : Physics
ALLEN
Illustration 13.
A car starts from rest with a constant tangential acceleration a0 in a circular path of radius r. At time t 0, the
car skids, find the value of coefficient of friction.
Solution
a0
The tangential and centripetal acceleration is provided only by the frictional force.
Thus, f sinq = ma0 and f cosq =
Þf=m
a 20 +
( a0 t0 )4 Þ ma
r
0
2
m ( a0 t0 )
mv 2
=
r
r
1+
2
q
f
2
v
R
a 20 t 04
= fmax
r2
a0
a2 t 4
1 + 02 0 .
g
r
Illustration 14.
LL
E
A small block slides with a velocity 0.5 gr on a horizontal frictionless A
surface as shown in the figure. The block leaves the surface at point C.
Calculate the angle q shown in the figure.
B
q r
C
O
r
20
Solution
v0
-2
Þ m=
19
a 20 t 40
r2
N
mmg = ma 0 1 +
0
(since the car skids beyound this speed)
2
æ 3ö
1
1
3
m (rgcosq) –
m 0.5 gr = mgr (1 - cos q) Þ cos q =
Þ q = cos–1 ç ÷
è 4ø
2
2
4
)
\\\\\\\\\\\\\\\\\\\\\\\
A rigid rod of length l and negligible mass has a ball of mass m attached to one end
with its other end fixed, to form a pendulum as shown in figure. The pendulum is inverted,
with the rod vertically up, and then released. Find the speed of the ball and the tension
in the rod at the lowest point of the trajectory of ball.
l
m
Solution
From COME : 2mgl =
1
mv 2 Þ v = 4gl = 2 gl
2
At the lowest point, laws of circular dynamics yield, T - mg =
52
mv 2
m
Þ T = mg + ( 4gl) = 5mg .
l
l
Z:\NODE02\B0AI-B0\TARGET\PHY\ENG\MODULE_02\02-CIRCULAR MOTION\01-THEORY.P65
A
Illustration 15.
(
Se
Þ
1
1
mvC2 –
mv02 = mgr (1-cosq)
2
2
ss
io
By energy conservation at points B & C,
mv 2C
r
n
As the block leaves the surface at C so there, normal reaction = 0 Þ mgcos q =
E
Pre-Medical : Physics
ALLEN
Illustration 16.
A particle of mass m tied to a string of length l and given a circular motion in the vertical plane. If it performs
the complete loop motion then prove that difference in tensions at the lowest and the highest point is 6 mg.
Solution
Let the speeds at the lowest and highest points be u and v respectively.
mu 2
l
...(i)
mv 2
- mg .
l
..(ii)
At the lowest point, tension = TL = mg +
At the highest point, tension = TH =
By conservation of mechanical energy,
\ From eqn. (ii) & (iii)
b g Þ
m v 2 + 4gl
l
TL – TH = 6mg
Illustration 17.
u2 = v2 + 4gl
...(iii)
N
Substituting this in eqn (i) TL = mg +
mu2 mv 2
= mg 2l
2
2
H
(i)
19
the tension in the string when it makes an angle q2 with the vertical.
h = l(cosq2 – cosq1)
ss
io
1
mv 2 = mgh Þ v= 2gh = 2gl(cos q2 - cos q1 )
2
Z:\NODE02\B0AI-B0\TARGET\PHY\ENG\MODULE_02\02-CIRCULAR MOTION\01-THEORY.P65
E
mv 2
where v =
l
lcosq1
lcosq2
2gl(cos q2 - cos q1 )
Se
A
At position B, T – mgcosq2 =
Þ T = mgcosq2 +
A
[AIPMT (Mains) 2008]
Applying conservation of mechanical energy between points A & B
(ii)
q1
20
Solution
speed of the particle and
n
(ii)
LL
E
q1 with the vertical and is released then obtain expressions for the :
(i)
q2
-2
it behaves as a simple pendulum. Now the string is pulled to point A making an angle
0
A particle of mass m is connected to a light inextensible string of length l such that
q2
l
q1
A
T
h
B
mg sin q2
q2
mg mg cos q2
m
[2gl(cosq2 – cosq1)] = mg(3cosq2 – 2cosq1).
l
BEGINNER'S BOX-2
1.
A particle of mass m1 is fastened to one end of a string and another one of mass m2 to the middle point;
the other end of the string being fastened to a fixed point on a smooth horizontal table. The particles are
then projected, so that the two portions of the string are always in the same straight line and describe horizontal
circles. Find the ratio of the tensions in the two parts of the string.
2.
A road is 8 m wide. Its average radius of curvature is 40 m. The outer edge is above the lower edge by
a distance of 1.28 m. Find the velocity of vehicle for which the road is most suited ? (g = 10 m/s2)
3.
A stone of mass 1 kg tied to a light string of length l = 10 m is whirling in a circular path in the vertical
plane. If the ratio of the maximum to minimum tensions in the string is 3, find the speeds of the stone at
the lowest and highest points.
53
Pre-Medical : Physics
Speed at D
(b)
Normal reaction at D
(c)
Height H
D
H
B
vC = 7gR
C
R
A car is moving along a hilly road as shown (side view). The coefficient of static friction between the tyres and
the pavement is constant and the car maintains a steady speed. If at one of the points shown the driver applies
brakes as hard as possible without making the tyres slip, the magnitude of the frictional force immediately
after the brakes are applied will be maximum if the car was at :(A)
point A
(B)
point B
(C)
point C
(D)
friction force same for positions A, B and C
C
A
B
-2
0
A stone weighing 0.5 kg tied to a rope of length 0.5 m revolves along a circular path in a vertical plane.
The tension of the rope at the bottom point of the circle is 45 newton. To what height will the stone rise
if the rope breaks at the moment when the velocity is directed upwards? (g=10 m/s2)
ss
io
n
20
LL
E
6.
(a)
A
19
5.
Calculate the following for the situation shown :-
N
4.
ALLEN
1. (C)
2. (D)
3. 6 m/s2 towards the centre.
4. 1 : 1
5. (C)
6. (C)
7. (A)
8. -3kˆ rad/s2 , -2kˆ rad/s
54
1.
BEGINNER'S BOX-2
2m1
m2 + 2m1
2. 8 m/s
3. vlowest = 20 2 m/s; vhighest = 20 m/s
4.
(a)
5. (B)
6. 1.5 m
5gR (b) 4mg (c)
9
R
2
Z:\NODE02\B0AI-B0\TARGET\PHY\ENG\MODULE_02\02-CIRCULAR MOTION\01-THEORY.P65
A
BEGINNER'S BOX-1
Se
ANSWER KEYS
E
Pre-Medical : Physics
ALLEN
Build Up Your Understanding
EXERCISE-I (Conceptual Questions)
KINEMATICS OF CIRCULAR MOTION
A particle of mass 'm' describes a circle of radius
(r). The centripetal acceleration of the particle is
4
. The momentum of the particle :–
r2
(4)
(4) changes in magnitude and direction
4m
7.
r
A particle is moving around a circular path with
uniform angular speed (w). The radius of the circular
path is (r). The acceleration of the particle is :–
w
r
2
(1)
(3) vw
(2)
w
r
8.
(4) vr
A car moves on a circular road, describing equal
angles about the centre in equal intervals of times.
Which of the statements about the velocity of car
is true :–
(4) 20 r.p.m.
A particle moving along a circular path. The angular
velocity, linear velocity, angular acceleration and
centripetal acceleration of the particle at any instant
r , r , r , r . Which of the
respectively are w
v a
ac
®
®
®
(3) a,b,c
(4) a,c,d
ss
io
9.
A particle is acted upon by a force of constant
magnitude which is always perpendicular to the
velocity of the particle. The motion of the particle
takes place in a plane. It follows, that :–
(2) 5.3 cm/s
(4) it moves in a straight line
(3) 0.44 cm/s
(4) none of these
A
(1) 2.3 cm/s
A particle moves in a circle of the radius 25 cm at
two revolutions per second. The acceleration of the
particle in m/sec2 is :–
(3) 4p2
(4) 2p2
®
(2) b,c,d
(1) its velocity is constant
(2) 8p2
®
(d) w ^ a c
An insect trapped in a circular groove of radius
12 cm moves along the groove steadily and
completes 7 revolutions in 100 s. What is the linear
speed of the motion :–
(1) p2
®
(c) v ^ a c
n
(1) a,b,d
®
(b) w ^ a
Se
(4) velocity is directed towards the centre of circle
Z:\NODE02\B0AI-B0\TARGET\PHY\ENG\MODULE_02\02-CIRCULAR MOTION\02-EXERCISE..P65
(3) 10 r.p.m.
®
(3) both magnitude and direction of velocity
change
E
(2) 14 r.p.m.
(a) w ^ v
(2) magnitude of velocity is constant but the
direction of velocity change
5.
(1) 7 r.p.m.
following relation is/are correct :–
(1) velocity is constant
4.
A mass of 2 kg is whirled in a horizontal circle by
means of a string at an initial speed of 5 r.p.m.
Keeping the radius constant the tension in the
string is doubled. The new speed is nearly :–
LL
E
3.
r
0
4m
r
(3) change in direction
2m
-2
2.
(2)
(2) change in magnitude
19
(3)
2m
r
(1) remains constant
N
(1)
A particle moves in a circle describing equal angle
in equal times, its velocity vector :–
20
1.
6.
(2) its K.E. is constant
(3) its acceleration is constant
10.
If the equation for the displacement of a particle
moving on a circular path is given by
(q) = 2t 3 + 0.5, where q is in radians and t in
seconds, then the angular velocity of the particle
after 2 s from its start is :–
(1) 8 rad/s
(2) 12 rad/s
(3) 24 rad/s
(4) 36 rad/s
55
Pre-Medical : Physics
(2) Constant
(3) Zero
(4) None of the above
A particle of mass (m) revolving in horizontal circle
of radius (R) with uniform speed v. When particle
goes from one end to other end of diameter, then
:-
18.
1
mv 2
2
19.
r
R
(2)
R
r
(4) change in momentum is 2 mv
(3) 860 cm/s2
(4) 990 cm/s2
A
(2) 720 cm/s2
(4)
R
r
Angular velocity of minute hand of a clock is :–
(1)
p
rad/s
30
(2) 8p rad/s
(3)
2p
rad/s
1800
(4)
p
rad/s
1800
A car moving with speed 30 m/s on a circular path
of radius 500 m. Its speed is increasing at the rate
of 2m/s2. The acceleration of the car is :–
(2) 1.8 m/s2
(3) 2 m/s2
(4) 2.7 m/s2
If a particle is rotating uniformly in a horizontal circle,
then –
ss
io
20.
(1) 9.8 m/s2
(1) no force is acting on particle
(2) velocity of particle is constant
Se
A stone is tied to one end of string 50 cm long and
is whirled in a horizontal circle with constant speed.
If the stone makes 10 revolutions in 20 s, then
what is the magnitude of acceleration of the stone
:(1) 493 cm/s2
(3) 1
20
(3) no change in momentum
For a particle in a non-uniform accelerated circular
motion :–
(1) velocity is radial and acceleration is transverse
only
(2) velocity is transverse and acceleration is radial
only
(3) velocity is radial and acceleration has both radial
and transverse components
(4) velocity is transverse and acceleration has both
radial and transverse components
56
Two particles having mass 'M' and 'm' are moving in
a circular path having radius R and r. If their time
period are same then the ratio of angular velocity
will be :–
(1)
(2) K.E. change by mv2
15.
(4) 60 degree / s
(3) particle has no acceleration
(4) no work is done
21.
A body of mass 1 kg tied to one end of string is
revolved in a horizontal circle of radius 0.1 m with
a speed of 3 revolution/sec, assuming the effect of
gravity is negligible, then linear velocity, acceleration
and tension in the string will be :–
(1) 1.88 m/s, 35.5 m/s2, 35.5 N
(2) 2.88 m/s, 45.5 m/s2, 45.5 N
(3) 3.88 m/s, 55.5 m/s2, 55.5 N
(4) None of these
Z:\NODE02\B0AI-B0\TARGET\PHY\ENG\MODULE_02\02-CIRCULAR MOTION\02-EXERCISE..P65
(1) K.E. changes by
14.
(3) 10.47 degree / s
-2
(1) rw2
(2) 10.47 rad / s
0
A body moves with constant angular velocity on a
circle. Magnitude of angular acceleration :-
(1) 1.66 rad / s
19
(3) mg.2pl
æ mv 2 ö
(4) çç l ÷÷ .l
è
ø
17.
ALLEN
The angular velocity of a particle rotating in a circular
orbit 100 times per minute is
N
13.
(1) 0
æ mv 2 ö
(2) çç l ÷÷ .2pl
è
ø
16.
n
12.
A sphere of mass m is tied to end of string of length
l and rotated through the other end along a
horizontal circular path with speed v. The work
done in full horizontal circle is :-
LL
E
11.
E
Pre-Medical : Physics
ALLEN
20
)m
p
with constant tangential acceleration. If the velocity
A particle moves along a circle of radius (
of the particle is 80 m/s at the end of the second
revolution after motion has begun, the tangential
acceleration is :–
(1) uniform but non zero
(2) zero
(3) variable
(4) 40 p m/s
2
(4) as can not be predicted from given information
The angular velocity of a wheel is 70 rad/s. If the
radius of the wheel is 0.5 m, then linear velocity of
the wheel is :–
(4) 20 m/s
(1) Doubled of previous value
(2) Equal to previous value
(3) Triple of previous value
(4) One third of previous value
29.
(1) no acceleration is present in the body
(2) no force acts on the body
(1) p2 m/s2 and direction along the tangent to the
(3) its velocity remains constant
circle.
p2
m/s2 and direction along the radius towards
4
the centre.
A
from the centre.
Z:\NODE02\B0AI-B0\TARGET\PHY\ENG\MODULE_02\02-CIRCULAR MOTION\02-EXERCISE..P65
25.
E
26.
(1) Forward
(2) Backward
(3) Towards the centre of the path
(4) Away from the centre of the path
31.
A fly wheel rotating at 600 rev/min is brought
under uniform deceleration and stopped after
2 minutes, then what is angular deceleration in
rad/sec2 ?
p
(1)
6
(2) 10 p
1
(3)
12
(4) 300
The linear and angular acceleration of a particle
are 10 m/s2 and 5 rad/s2 respectively. It will be
at a distance from the axis of rotation.
(1) 50 m
1
(2)
m
2
(3) 1 m
A pendulum is suspended from the roof of a rail
road car. When the car is moving on a circular track
the pendulum inclines :
Se
(4) p2 m/s2 and direction along the radius away
30.
n
the centre.
(4) no work gets done on it
ss
io
(2) p2 m/s2 and direction along the radius towards
(3)
When a body moves with a constant speed along
a circle :–
LL
E
A stone tied to the end of a string of 1m long is
whirled in a horizontal circle with a constant speed.
If the stone makes 22 revolution in 44 seconds,
what is the magnitude and direction of acceleration
of the stone :–
0
(3) 30 m/s
If the speed and radius both are trippled for a body
moving on a circular path, then the new centripetal
force will be :–
-2
(2) 35 m/s
28.
19
(1) 70 m/s
The angular acceleration of particle moving along
a circular path with uniform speed :–
N
(3) 160 p m/s
2
24.
27.
(2) 640 p m/s2
(1) 40 m/s2
23.
DYNAMICS OF HORIZONTAL CIRCULAR
MOTION
20
22.
(4) 2 m
32.
A string of length 0.1 m cannot bear a tension more
than 100N. It is tied to a body of mass 100g and
rotated in a horizontal circle. The maximum angular
velocity can be (1) 100 rad/s
(2) 1000 rad/s
(3) 10000 rad/s
(4) 0.1 rad/s
The radius of the circular path of a particle is
doubled but its frequency of rotation is kept
constant. If the initial centripetal force be F, then
the final value of centripetal force will be :–
(1) F
(2)
F
2
(3) 4F
(4) 2F
57
Pre-Medical : Physics
A 0.5 kg ball moves in a circle of radius 0.4 m
at a speed of 4 m/s. The centripetal force on the
ball is :–
(3) 40N
(1) 250 N
A body is revolving with a constant speed along a
circle. If its direction of motion is reversed but the
speed remains the same then :–
40.
(a) the centripetal force will not suffer any change
in magnitude
(c) the centripetal force will not suffer any change
in direction
(d) the centripetal force is doubled
(3) c,d
(4) a, c
a r and a t represent radial and tangential
acceleration. The motion of a particle will be
uniform circular motion, if :–
(4)
1
times
4
The force required to keep a body in uniform circular
motion is :(1) Centripetal force
(2) Centrifugal force
(3) Resistance
(4) None of the above
BANKING OF TRACKS
42.
A car moving on a horizontal road may be thrown
out of the road in taking a turn :–
(3) ar ¹ 0 but at = 0
(4) ar ¹ 0 and at ¹ 0
(2) due to lack of proper centripetal force
20
(1) by the gravitational force
In uniform circular motion, the velocity vector and
acceleration vector are
(3) due to rolling friction between the tyres and
the Road
ss
io
(4) due to reaction of the road
43.
(3) Opposite direction
(1) 20 rad/s
(2) 40 rad/s
(3) 100 rad/s
(4) 200 rad/s
The earth (Me = 6 × 1024 kg) is revolving round the
sun in an orbit of radius (1.5 ×108) km with angular
velocity of (2 × 10–7) rad/s. The force (in newton)
exerted on the earth by the sun will be :(1) 36 × 10
(3) 25 × 10
16
(2) 16 × 10
44.
v2R
(3)
bg
(4)
v2 b
R
A boy holds a pendulum in his hand while standing
at the edge of a circular platform of radius r rotating
at an angular speed w. The pendulum will hang at
an angle q with the vertical so that :–
(Neglect length of pendulum)
(1) tan q = 0
(2) tan q =
w2 r 2
g
rw2
g
(4) tan q =
g
w2 r
24
(4) Zero
v
(2)
Rgb
v2 b
(1)
Rg
(3) tan q =
Z:\NODE02\B0AI-B0\TARGET\PHY\ENG\MODULE_02\02-CIRCULAR MOTION\02-EXERCISE..P65
A
A string of length 10 cm breaks if its tension exceeds
10 newton. A stone of mass 250 g tied to this string,
is rotated in a horizontal circle. The maximum
angular velocity of rotation can be :-
21
Radius of the curved road on national highway is
R. Width of the road is b. The outer edge of the
road is raised by h with respect to inner edge so
that a car with velocity v can pass safely over it.
The value of h is :–
Se
(4) Not related to each other
58
(3) 4 times
(2) ar = 0 but at ¹ 0
(2) Same direction
38.
(2) half
(1) ar = 0 and at = 0
(1) Perpendicular to each other
37.
41.
(1) double
n
36.
(4) 1200 N
A motor cycle driver doubles its velocity when he is
taking a turn. The force exerted towards the centre
will become :-
LL
E
35.
(2) b,c
(3) 750N
N
(b) the centripetal force will have its direction
reversed
(1) a,b
(2) 1000N
(4) 80N
0
34.
(2) 20N
ALLEN
A 500 kg car takes a round turn of radius 50 m with
a velocity of 36 km/hr. The centripetal force is :-
-2
(1) 10N
39.
19
33.
E
Pre-Medical : Physics
A stone attached to one end of a string is whirled
in a vertical circle. The tension in the string is
maximum when :–
(1) the string is horizontal
(2) the string is vertical with the stone at highest
position
(3) the string is vertical with the stone at the lowest
position
(4) the string makes an angle of 45° with the vertical
51.
A weightless thread can withstand tension upto
30 N. A stone of mass 0.5 kg is tied to it and is
revolved in a circular path of radius 2m in a vertical
plane. If g = 10 m/s2, then the maximum angular
velocity of the stone can be :–
Let 'q' denote the angular displacement of a simple
pendulum oscillating in a vertical plane. If the mass
of the bob is (m), then the tension in string is mg
cosq :–
(1) always
(2) never
(4) at the mean position
A pendulum bob has a speed 3 m/s while passing
through its lowest position, length of the pendulum
is 0.5 m then its speed when it makes an angle
of 60o with the vertical is :–( g = 10 m/s2)
(1) 2 m/s
(3) 4 m/s
(4) 3 m/s
The mass of the bob of a simple pendulum of length
L is m. If the bob is left from its horizontal position
then the speed of the bob and the tension in the
thread in the lowest position of the bob will be
respectively :–
(1)
(3)
52.
60 rad/s
2gL and 3 mg
L
(3) 2 mg and
(1) x = L
(2) x = 2L
O
m
L
2gL
(2) 16 N
(3) 5 N
(4) 10 N
54.
In a vertical circle of radius (r), at what point in its
path a particle may have tension equal to zero :–
(3) at any point
(4) at a point horizontal from the centre of radius
O
x
A
(1) T1 = T2
(2) T1 > T2
(3) T1 < T2
(4) T1 ³ T2
C
A body crosses the topmost point of a vertical circle
with critical speed. What will be its centripetal
acceleration when the string is horizontal :–
(1) g
(1) highest point
(2) lowest point
v
A particle is moving in a vertical circle the tension
in the string when passing through two position
at angle 30o and 60o from vertical from lowest
position are T1 and T2 respectively then :–
Se
A
Z:\NODE02\B0AI-B0\TARGET\PHY\ENG\MODULE_02\02-CIRCULAR MOTION\02-EXERCISE..P65
E
(1) 6 N
L
(4) x= 2L
53.
A stone of mass 1 kg is tied to the end of a string
of 1 m length. It is whirled in a vertical circle. If
the velocity of the stone at the top be 4 m/s. What
is the tension in the string (at that instant) ?
P
n
2gL
(4) 2 gL and 3 mg
49.
(4) 10 rad/s
ss
io
(2) 3 mg and
30 rad/s
A body tied to a string of length L is revolved in a
vertical circle with minimum velocity, when the body
reaches the upper most point the string breaks and
the body moves under the influence of the
gravitational field of earth along a parabolic path.
The horizontal range AC of the body will be :–
(3) x= 2 2L
48.
(2)
LL
E
47.
(2) 1 m/s
(1) 5 rad/s
N
46.
0
(3) at the extreme positions
20
45.
50.
-2
VERTICAL CIRCULAR MOTION
19
ALLEN
55.
(2) 2g
(3) 3g
(4) 6g
Stone tied at one end of light string is whirled round
a vertical circle. If the difference between the
maximum and minimum tension experienced by
the string wire is 2 kg wt, then the mass of the stone
must be :–
(1) 1 kg
(2) 6 kg
(3) 1/3 kg
(4) 2 kg
59
Pre-Medical : Physics
A mass tied to a string moves in a vertical circle with
a uniform speed of 5 m/s as shown. At the point
P the string breaks. The mass will reach a height
above P of nearly (g = 10 m/s2) :–
ALLEN
59.
(1) 1 m
(2) 0.5 m
O
m
P
(2) B
(3) 1.75 m
2
2
m v g
r
(4)
v g
r
A frictionless track ABCDE ends in a circular loop
of radius R. A body slides down the track from
point A which is at a height h = 5 cm. Maximum
value of R for the body to successfully complete
the loop is :–
61.
(1) 5 cm.
3gr
(2)
2gr
(3)
gr
(4)
gr
2
A stone of mass 0.2 kg is tied to one end of a thread
of length 0.1 m whirled in a vertical circle. When
the stone is at the lowest point of circle, tension in
thread is 52N, then velocity of the stone will be :–
(1) 4 m/s
(2) 2 cm.
62.
(3) 6 m/s
(4) 7 m/s
A suspended simple pendulum of length l is making
an angle q with the vertical. On releasing, its
velocity at lowest point will be :-
n
10
cm.
3
(2) 5 m/s
ss
io
(3)
(1)
LL
E
58.
mv 2
r
(1)
2gl(1 + cos q )
(2)
2gl sin q
(3)
2gl(1 - cos q)
(4)
2gl
Se
A
15
(4)
cm.
4
ANSWER KEY
EXERCISE-I (Conceptual Questions)
Que.
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
Ans.
2
3
2
2
3
3
1
4
2
3
1
3
4
1
4
Que.
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
Ans.
2
3
4
4
4
1
1
2
2
1
4
2
3
4
4
Que.
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
Ans.
1
4
2
4
3
1
1
1
2
3
1
2
1
3
3
Que.
Ans.
46
1
47
1
48
1
49
1
50
3
51
1
52
2
53
2
54
3
55
3
56
4
57
1
58
2
59
2
60
3
Que.
61
62
Ans.
2
3
60
Z:\NODE02\B0AI-B0\TARGET\PHY\ENG\MODULE_02\02-CIRCULAR MOTION\02-EXERCISE..P65
(3)
(2) mg -
-2
2
mv 2
r
A fighter plane is moving in a vertical circle of radius
'r'. Its minimum velocity at the highest point of the
circle will be :–
19
(1) mg +
60.
C
B
(4) D
N
If the overbridge is concave instead of being convex,
then the thrust on the road at the lowest position
will be :-
O
D
(3) C
(4) 1.25 m
57.
A
(1) A
0
1m
A particle of mass m is performing vertical circular
motion (see figure). If the average speed of the
particle is increased, then at which point maximum
breaking possibility of the string :–
20
56.
E
Pre-Medical : Physics
ALLEN
AIPMT/NEET & AIIMS (2006-2018)
EXERCISE-II (Previous Year Questions)
AIPMT (Pre) 2012
AIPMT 2006
A tube of length L is filled completely with an
6.
incompressible liquid of mass M and closed at both
the ends. The tube is then rotated in a horizontal
A car of mass 1000 kg negotiates a banked curve
of radius 90 m on a fictionless road. If the banking
angle is 45°, the speed of the car is :-
plane about one of its ends with a uniform angular
(1) 5 m/s
(2) 10 m/s
velocity w. The force exerted by the liquid at the
(3) 20 m/s
(4) 30 m/s
other end is :–
M L 2 w2
2
(3) M L w2 (4)
A car runs at a constant speed on a circular track
of radius 100 m, taking 62.8 seconds for every
circular lap. The average velocity and average speed
for each circular lap respectively is :(1) 0,0
(2) 0, 10 m/s
(3) 10 m/s, 10 m/s
(4) 10 m/s, 0
3.
8.
ms Rg
(3)
ms mRg
(4)
Rg / m s
Two stones of masses m and 2 m are whirled in
20
(1) 16 m/s and 17 m/s (2) 13 m/s and 14 m/s
(3) 3
AIPMT (Pre) 2010
A gramophone record is revolving with an angular
mg
w2
2
(3) r <
w
mg
(2) r = mgw2
mg
(4) r £ 2
w
AIPMT 2011
A particle moves in a circle of radius 5 cm with
constant speed and time period 0.2p s. The
acceleration of the particle is :(1) 15 m/s2
(2) 25 m/s2
(3) 36 m/s2
(4) 5 m/s2
9.
(2) 2
(4) 4
r
The position vector of a particle R as a function
of time is given by :r
R = 4 sin(2pt)iˆ + 4 cos(2pt)ˆj
Se
A
velocity w. A coin is placed at a distance r from the
centre of the record. The static coefficient of friction
is µ. The coin will revolve with the record if :(1) r ³
(1) 1
ss
io
(3) 14 m/s and 15 m/s (4) 15 m/s and 16 m/s
r
2
and the lighter one in radius r. The tangential speed
of lighter stone is n times that of the value of heavier
stone when they experience same centripetal
forces. The value of n is :
(g = 10 m/s2)
Z:\NODE02\B0AI-B0\TARGET\PHY\ENG\MODULE_02\02-CIRCULAR MOTION\02-EXERCISE..P65
(2)
horizontal circles, the heavier one in a radius
speed of the car at the top of the hill is between.
E
mRg / m s
A roller coaster is designed such that riders
of a hill whose radius of curvature is 20 m. The
5.
(1)
Re-AIPMT 2015
experiece"weightlessness" as they go round the top
4.
A car of mass m is moving on a level circular track
of radius R. If ms represents the static friction
between the road and tyres of the car, the maximum
speed of the car in circular motion is given by :-
LL
E
AIPMT 2008
7.
0
M L 2w
2
-2
(2)
19
2.
M L w2
2
N
(1)
AIPMT (Mains) 2012
n
1.
Where R is in meters, t is in seconds and î and
ĵ denote unit vectors along x and y-directions,
respectively. Which one of the following statements
is wrong for the motion of particle ?
(1) Path of the particle is a circle of radius 4 meter
r
(2) Acceleration vectors is along -R
(3) Magnitude of acceleration vector is
v2
where
R
v is the velocity of particle.
(4) Magnitude of the velocity of particle is
8 meter/second
61
Pre-Medical : Physics
ALLEN
NEET-I 2016
(4) Velocity is perpendicular to rr and acceleration
(2) 0.15 m/s2
is directed away from the origin
(3) 0.18 m/s2
(4) 0.2 m/s2
AIIMS 2016
(4)
5gR
A car is negotiating a curved road of radius R. The
road is banked at an angle q. the coefficient of
friction between the tyres of the car and the road
is µ s. The maximum safe velocity on this road is :µ s + tan q
1 - m s tan q
16.
(2)
gR
(4)
g µ s + tan q
R 2 1 -m s tan q
NEET-II 2016
In the given figure, a = 15 m/s2 represents the total
acceleration of a particle moving in the clockwise
direction in a circle of radius R = 2.5 m at a given
instant of time. The speed of the particle is :-
(1) T +
mv 2
l
(4) T
NEET(UG) 2018
A body initially at rest and sliding along a frictionless
track from a height h (as shown in the figure) just
completes a vertical circle of diameter AB = D. The
height h is equal to :-
Se
°
R
30
(2) T -
(3) Zero
17.
a
A
O
mv 2
l
(1) 5.7 m/s
(2) 6.2 m/s
(3) 4.5 m/s
(4) 5.0 m/s
B
h
A particle moves so that its position vector is given
r
by r = cos wt xˆ + sin wt yˆ . Where w is a constant.
A
(1)
Which of the following is true ?
3
D
2
(2) D
(3)
7
D
5
(4)
5
D
4
ANSWER KEY
Que.
Ans.
Que.
Ans.
62
1
1
2
2
16
4
17
4
3
3
4
4
5
4
6
4
7
2
8
2
9
4
10
1
11
4
12
2
13
1
14
3
15
1
Z:\NODE02\B0AI-B0\TARGET\PHY\ENG\MODULE_02\02-CIRCULAR MOTION\02-EXERCISE..P65
g µ s + tan q
R 1 -ms tan q
(3)
14.
One end of string of length l is connected to a
particle of mass 'm' and the other end is connected
to a small peg on a smooth horizontal table. If the
particle moves in circle with speed 'v' the net force
on the particle (directed towards centre) will be
(T represents the tension in the string) :-
20
gR 2
(4) 6 sec
NEET(UG) 2017
LL
E
µ s + tan q
1 - m s tan q
(1) 7.5 sec (2) 4.5 sec (3) 5 sec
0
3gR
-2
(3)
2gR
A particle starts with angular acceleration
2 rad/sec2. It moves 100 rad in a random interval
of 5 sec. Find out the time at which random intervel
starts.
19
(2)
gR
15.
N
What is the minimum velocity with which a body
of mass m must enter a vertical loop of radius R
so that it can complete the loop ?
(1)
13.
is directed towards the origin
(1) 0.1 m/s2
(1)
12.
(3) Velocity is perpendicular to rr and acceleration
n
11.
A particle of mass 10 g moves along a circle of
radius 6.4 cm with a const ant tangential
acceleration. What is the magnitude of this
acceleration if the kinetic energy of the particle
becomes equal to 8 × 10–4 J by the end of the
second revolution after the beginning of the
motion ?
ss
io
10.
(1) Velocity and acceleration both are perpendicular
r
to r .
(2) Velocity and acceleration both are parallel to rr
E
Pre-Medical : Physics
ALLEN
Check Your Understanding
EXERCISE-III (Analytical Questions)
(2) 9 sec.
(3) 4 sec.
(4) 10 sec.
l. The mass goes around a verticle circular path
with the other end hinged at the centre. What should
be the minimum velocity of mass at the bottom of
the circle so that the mass completes the circle ?
Keeping the banking angle of the road constant, the
maximum safe speed of passing vehicles is to be
increased by 10%. The radius of curvature of the
5.
(4) 30.5 m
Three identical particles are joined together by a
thread as shown in figure. All the three particles are
moving in a horizontal plane. If the velocity of the
outermost particle is v0, then the ratio of tensions in
the three sections of the string is :-
(2)
3gl
(3)
5gl
(4)
gl
A stone is tied to a string of length ‘ l’ and is whirled
N
(3) 24.20 m
4gl
(1)
u2 - gl
(3)
2gl
LL
E
3.
(2) 18 m
(1)
in a vertical circle with the other end of the string
as the centre. At a certain instant of time, the stone
is at its lowest position and has a speed ‘u’. The
magnitude of the change in velocity as it reaches a
position where the string is horizontal (g being acceleration due to gravity) is :–
road will have to be changed from 20 m to :(1) 16 m
A mass m is attached to the end of a rod of length
(3) 7 : 11 : 6
O
l
A
l
B
l
u2 - 2gl
(4)
2(u 2 - gl)
C
n
(2) 3 : 4 : 5
(2) u –
20
(1) 3 : 5 : 7
0
(1) 7 sec.
4.
-2
2.
A wheel has a constant angular acceleration of
3.0 rad/s2, During a certain 4.0 s interval, it turns
through an angle of 120 rad. Assuming that at
t = 0, angular speed w0 = 3 rad/s how long, is
motion at the start of this 4.0 second interval ?
19
1.
Z:\NODE02\B0AI-B0\TARGET\PHY\ENG\MODULE_02\02-CIRCULAR MOTION\02-EXERCISE..P65
A
Se
ss
io
(4) 3 : 5 : 6
E
EXERCISE-III (Analytical Questions)
Que.
Ans.
1
1
ANSWER KEY
2
3
3
4
4
1
5
4
63
Pre-Medical : Physics
ALLEN
Target AIIMS
EXERCISE-IV (Assertion & Reason)
Directions for Assertion & Reason questions
These questions consist of two statements each, printed as Assertion and Reason. While answering
these Questions you are required to choose any one of the following four responses.
(A)
If both Assertion & Reason are True & the Reason is a correct explanation of the Assertion.
(B)
If both Assertion & Reason are True but Reason is not a correct explanation of the Assertion.
(C)
If Assertion is True but the Reason is False.
(D)
If both Assertion & Reason are false.
1.
Assertion :In uniform circular motion, the linear
speed and angular speed of the body are constant.
Reason : A body can move on a circular path
without having acceleration.
(1) A
(2) B
(3) C
(4) D
Assertion :The resultant acceleration of an object
in circular motion is towards the centre if the speed
is constant.
Reason : A vector is necessarily changed if it is
rotated through an angle.
(1) A
(2) B
(3) C
(4) D
Assertion : Work done by the centripetal force
in moving a body along a circle is always zero.
Reason : In circular motion the displacement of
the body is along the force.
(1) A
(2) B
(3) C
(4) D
Assertion :Centripetal and centrifugal forces
cancel each other.
Reason : This is because they are always equal
and opposite.
(1) A
(2) B
(3) C
(4) D
Assertion : Work done by centripetal force is zero.
Reason : Centripetal force acts perpendicular
to the displacement.
(1) A
(2) B
(3) C
(4) D
Assertion :When a particle moves in a circle with
a uniform speed its acceleration is constant but the
velocity changes.
Reason : Angular displacement is not an axial
vector.
(1) A
(2) B
(3) C
(4) D
Assertion :If a particle moves along circular path
with constant speed then acceleration must be
present.
Reason : If a particle moves with variable velocity
then acceleration must be present.
(1) A
(2) B
(3) C
(4) D
Assertion :In uniform circular motion speed of
particle must be constant.
Reason : In uniform circular motion no force
or acceleration is acting on particle acting parallel
or in anti parallel to the direction of velocity
(1) A
(2) B
(3) C
(4) D
7.
8.
0
-2
20
n
13.
14.
15.
ANSWER KEY
EXERCISE-IV (Assertion & Reason)
Que.
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
Ans.
3
3
3
4
1
4
1
1
1
1
4
3
1
3
1
64
Z:\NODE02\B0AI-B0\TARGET\PHY\ENG\MODULE_02\02-CIRCULAR MOTION\02-EXERCISE..P65
A
6.
12.
ss
io
5.
11.
Se
4.
19
10.
LL
E
3.
Assertion :When the direction of motion of a
particle moving in a circular path is reversed the
direction of radial acceleration still remains the same
(at the given point).
Reason : For a particle revolving on circular path
in any direction such as clockwise or anticlockwise,
the direction of radial acceleration is always towards
the centre of the circular path.
(1) A
(2) B
(3) C
(4) D
Assertion :For uniform circular motion it is
necessary that the speed of the particle is constant.
Reason : There is no tangential force or
tangential acceleration acting on particle in uniform
circular motion.
(1) A
(2) B
(3) C
(4) D
Assertion :Acceleration of the particle in uniform
circular motion remains constant.
Reason : Velocity of the particle doesn't change
in circular motion.
(1) A
(2) B
(3) C
(4) D
Assertion :In the inertial frame centrifugal force
can't appear.
Reason : In the uniform circular motion centripetal
force will counter balance to the centrifugal force.
(1) A
(2) B
(3) C
(4) D
Assertion :During a turn, the value of centripetal
force should be less than the limiting frictional
force.
Reason : The centripetal force is provided by the
frictional force between the tyres and the road.
(1) A
(2) B
(3) C
(4) D
Assertion :A car moving on a horizontal road
may be thrown out of the road in taking a turn due
to lack of proper centripetal force.
Reason : If a particle moves in a circle, describing
equal angles in equal intervals of time. Then the
velocity vector changes its magnitude.
(1) A
(2) B
(3) C
(4) D
Assertion :- When a car turns around a circular
path it is acted upon by a centripetal force.
Reason :- The friction acting on the wheels of car
provides necessary centripetal force.
(1) A
(2) B
(3) C
(4) D
N
2.
9.
E